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Answer 1 · 2018-02-02T19:31:20

#y = C_1 e^x + C_4 cos(2x) + C_5 sin(2x)#

The solution for this kind of differential equation (linear homogeneous with constant parameters) is

#y = C e^(lambda x)# and afetr substitution

#C(lambda^3-lambda^2+4lambda-4)e^(lambda x)=0#

then solving

#lambda^3-lambda^2+4lambda-4=0 rArr (lambda^2+4)(lambda-1)=0#

and the solution is

#y = C_1 e^x+C_2 e^(i 2x)+C_3 e^(-i 2 x)# now using the de Moivre's identity

#e^(i phi) = cos(phi)+i sin(phi)# and knowing that #y# is real we easily obtain

#y = C_1 e^x + C_4 cos(2x) + C_5 sin(2x)#